问题描述
我没有足够的声誉积分可以发表评论，但是无数次有人(错误地)建议使用log10来计算正整数的位数.这对于大数目是错误的！
I don't have enough reputation points yet to leave comments, but saw numerous times when people (incorrectly) suggest using log10 to calculate the number of digits in a positive integer. This is wrong for large numbers!
long n = 99999999999999999L;
// correct answer: 17
int numberOfDigits = String.valueOf(n).length();
// incorrect answer: 18
int wrongNumberOfDigits = (int) (Math.log10(n) + 1);
// also incorrect:
double wrongNumberOfDigits2 = Math.floor(Math.log10(n) + 1);
基于对数的解决方案将错误地输出18而不是17.
The logarithm-based solutions will incorrectly output 18 instead of 17.
我想了解原因.
Way to get number of digits in an int?
Fastest way to get number of digits on a number?
推荐答案
问题在于，在这种情况下，99999999999999999无法精确表示为(双精度)浮点值.当作为double参数传递给log10时，最接近的值为1.0E+17.
The problem is that 99999999999999999 cannot be exactly represented as a (double precision) floating-point value in this case. The nearest value is 1.0E+17 when passed as a double parameter to log10.
log10(n)值也是如此:16.999999999999999995657...-可以表示的最接近值是17.
The same would be true of the log10(n) value: 16.999999999999999995657... - the nearest value that can be represented is 17.
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